How do you integrate $int x^3e^(x^2-1)dx$ using integration by parts?

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Answer 1 · 2015-11-09T15:51:18

See the explanation.

$int x^3e^(x^2-1)dx$

Integrateing $x^3$ and differentiating $e^(x^2-1)$ would give us MORE $x$ 's, so let's try the other way around first.

To integrate $e^(x^2-1)$ , we'll need another $x$ so that we can substitute.

So,
$int x^3e^(x^2-1)dx = int x^2[xe^(x^2-1)dx]$

Let $u = x^2$ and $dv = xe^(x^2-1)dx$ .

We get $du = 2x dx$ and $v = 1/2e^(x^2-1)$ .

Our integral becomes

$= 1/2x^2e^(x^2-1) - intxe^(x^2-1)dx$ .

The integral may again be evaluated by substitution.

$= 1/2x^2e^(x^2-1) - 1/2e^(x^2-1) +C$ .

How do you integrate int x^3e^(x^2-1)dxint x^3e^(x^2-1)dx using integration by parts?